
Re: Fractious Fractions
Posted:
Jun 5, 2013 12:20 PM


On Tue, 4 Jun 2013 00:11:59 0700, William Elliot <marsh@panix.com> wrote:
>On Mon, 3 Jun 2013, dullrich@sprynet.com wrote: > >> >Let a = (aj)_j be a sequence of positive integers. >> > >> >Define [a1] = a1; [a1,a2] = a1 + 1/a2; [a1,a2,a3] = a1 + 1/(a2 + 1/a3); >> > [a1,.. aj] = a1 + 1/(a2 + 1/(a3 +..+ 1/aj))..) >> > >> >Shome use the more visual notation >> > a1 + 1/a2+ 1/a3+ ..+1/aj >> >for [a1,.. aj]. >> > >> >Let cj = [a1,.. aj]. >> > >> >Assume j < k < n. How does one show >> > ck  c_n < cj  c_n? >> >> It's enough to show this: >> >> (i) The numbers c_{j+1}  c_j are alternating >> in sign and decreasing in absolute value. > >Done using 2x2 matrices. In fact, c_(j+1) = cj  (1)^j / sj.s_(j+1) >where rj, sj are positive coprime integers with cj = rj/sj >and the r and s sequences are increasing. > >> (You have to figure out why what you want follows from (i)). > >That, in a nutshell, is my question. > >> Now (i) follows from (ii): >> >> (ii) The numbers c_{j+1}  c_j alternate in >> sign. If j is odd then c_{j+2} > c_j, >> while if j is even then c_{j+2} < c_j. > >Additionally, if j odd and k even, then cj < ck. > >> To prove (ii): Let d_j = [a_2,...,a_j], so that >> >> (*) c_j = 1 + 1/d_j. > >c2 = a1 + 1/d2. What did you intend?
c_j = a_1 + 1/d_j .
>> By induction you can assume (iii): >> >> (iii) The numbers d_{j+1}  d_j alternate in >> sign. If j is odd then d_{j+2} < d_j, >> while if j is even then d_{j+2} > d_j. >> >> And then it's not hard to show that (ii) >> follows from (iii) and (*). >>

